## Trigonometry (11th Edition) Clone

$sec~(cos^{-1}~\frac{u}{\sqrt{u^2+5}}) = \frac{\sqrt{u^2+5}}{u}$
Let $~~\theta = cos^{-1}~\frac{u}{\sqrt{u^2+5}}$ Then $~~cos~\theta = \frac{u}{\sqrt{u^2+5}}$ We can find an expression for $sec~\theta$: $sec~\theta = \frac{1}{cos~\theta}$ $sec~\theta = \frac{\sqrt{u^2+5}}{u}$ Therefore, $~~sec~(cos^{-1}~\frac{u}{\sqrt{u^2+5}}) = \frac{\sqrt{u^2+5}}{u}$